Suppose $M$ is a manifold. Morse function $f:M \rightarrow \mathbb{R}$ is defined as a function in which all its critical points are non-degenerate.

In the link provided, $f(x)=x^3$ is not a morse function because it has only one critical point, that is $x=0$. But this point is degenerate, so the function is not a Morse function.

My question is: Why do we need to include the condition 'non-degenerate' in the definition of Morse function?

Also in the link provided, there stated 'by rotating the coordinate system under the graph, the degenerate critical point either is removed or breaks up into two non-degenerate critical points.' Can anyone explain this statement to me? I don't understand what is meant by 'rotating the coordinate system under the graph'.


1 Answer 1


Morse functions are much nicer than any old function. Asking why they are defined as they are is equivalent to asking why continuous functions are required to be continuous.

Anyway one reason Morse functions are nice is that critical points are isolated and stable. A small perturbation of a Morse function has the same number of critical points, which are located at almost the same places. This is basically due to the implicit function theorem.

Moreover Morse functions help to understand the topology of a manifold. It is a fact that sublevel sets $f_c=\{x\in M\,|\, f(x)\leq c\}$ of a function $f:M\rightarrow \mathbb{R}$ only change when $c$ passes a critical value (there is a critical point $x$ such that $c=f(x)$). The changes in topology can be quite complicated, but for Morse functions they are well understood. If $c$ is a critical value of a Morse function $f$ such that there is only one critical point with that value than $f_{c-\epsilon}$ and $f_{c+\epsilon}$ differ by the attachment of a $k$-handle. Here $k$ is the index of the critical point, which captures its type. One can now understand the topology of a manifold by choosing a Morse function on it (they always exist in abundance) and tracking all the changes in topology due to the critical points.

I don't think that rotating the graph is a nice description what happens with a degenerate critical point. Better to think of it as follows:

Let $f(x)=x^3-\epsilon x$. For $\epsilon=0$ there is a degenerate critical point at the origin. If $\epsilon>0$ however, there are two critical points! For $\epsilon<0$ there is none. By the way, choosing $\epsilon x$ as a perturbation is irrelevant, I can choose almost any small function!


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